$\sqrt[8]{80}, \sqrt[4]{40}$
$={80}^{\frac{1}{8}}, {40}^{\frac{1}{4}}$
$={80}^{\frac{1}{8}}, {40}^{\frac{2}{8}}$
$={80}^{\frac{1}{8}}, {1600}^{\frac{1}{8}}$
Now,
   $80<1600$
$=>{80}^{\frac{1}{8}}<{1600}^{\frac{1}{8}}$
$=>\sqrt[8]{80}< \sqrt[4]{40}$

Rationalizing factor of  $\displaystyle \frac{2\sqrt{3}}{\sqrt{5}}$ is $\sqrt{5}$

$\dfrac{3\sqrt{3}}{2\sqrt{2}}=\dfrac{3\times 1.73214}{2\times 1.41429} = \dfrac{5.19642}{2.85828}
=1.82151$
$\dfrac{2\sqrt{2}}{3\sqrt{3}}=\dfrac{2\times 1.41429}{3\times 1.73214}=\dfrac{2.85828}{5.19642}=0.55004$
$\therefore \dfrac{3\sqrt{3}}{2\sqrt{2}} >\dfrac{2\sqrt{2}}{3\sqrt{3}}$

$\sqrt 5$ > $\sqrt 4$ > $\sqrt 3$ > $\sqrt 2$

$\displaystyle \therefore $ $\displaystyle \sqrt[6]{3}$ = $\displaystyle \left ( 3 \right )^{\dfrac{1}{6}}$ = $\displaystyle \left ( 3^{2} \right )^{\dfrac{1}{12}}$ = $\displaystyle \left ( 9 \right )^{\dfrac{1}{12}}$
$\displaystyle \sqrt{2}$ = $\displaystyle \left ( 2 \right )^{\dfrac{1}{2}}$ = $\displaystyle \left ( 2^{6} \right )^{\dfrac{1}{12}}$ = $\displaystyle \left ( 64 \right )^{\dfrac{1}{12}}$
$\displaystyle \sqrt[3]{4}$ = $\displaystyle \left ( 4 \right )^{\dfrac{1}{3}}$ = $\displaystyle \left ( 4^{4} \right )^{\dfrac{1}{12}}$ = $\displaystyle \left ( 256 \right )^{\dfrac{1}{12}}$

Since, fifth root of four is smallest.

$p= \sqrt{32} - \sqrt{24}$

$p= 5.656-4.898 = 0.758$

$q = \sqrt{50} + \sqrt{48}$

$q= 7.071- 6.928 = 0.089$

$\therefore p> q$

$\sqrt{2} =1.414$

$\sqrt{17} =4.123$

Hence,

$x = \sqrt{2}+1 =1.414+1 = 2.414$

$ y= \sqrt{17} -\sqrt{2} =4.123- 1.414 = 2.709$

Hence , $x<y$

LCM of $3, 6, 4 = 12$

$x = \sqrt{7}-\sqrt{5}$

$= 2.64 – 2.23= 0.41$

$y=  \sqrt{13}-\sqrt{11}$

$= 3.60- 3.31 = 0.29$

$Hence, x> y$

$\sqrt{17} \approx 4.123$

Yes.

Let x $= 21, y =\sqrt{2}$ be a rational number
Now $x+y=21 +\sqrt{2}=21+1.4142....=22.4142....$ , which is non-terminating and non-recurring. Hence x+y is irrational.

$\sqrt[3]{2} \approx 1.26$

Clearly, one of $4^{\frac{1}{5}}$ and $5^{\frac{1}{4}}$ must be the smallest.
$\frac{1}{5}$<$\frac{1}{4}$.
So, $\displaystyle 4^{\frac{4}{20}}$ < $5^{\frac{5}{20}}$

$12<13<15<17\ \Rightarrow \sqrt{12}<\sqrt{13}<\sqrt{15}<\sqrt{17}$

$\sqrt[4]{64}=\sqrt[4]{2^6}=2\sqrt[4]{2^2}=2\sqrt{2}=2\sqrt[6]{2^3}=2\sqrt[6]{8}$
$\sqrt[6]{128}=\sqrt[6]{2^7}=2\sqrt[6]{2}$
$\sqrt[4]{64}>\sqrt[6]{128}$

(B) $\sqrt[3]{4}, \sqrt[4]{5},  \sqrt[4]{6}, \sqrt[3]{8}$

$=4^{1/3}, 5^{1/4}, 6^{1/4}, 8^{1/3}$

L.C.M of 3 & 4 $=12$

So, the given surds can be written as,

$=4^{4/12}, 5^{3/12}, 6^{3/12}, 8^{4/12}$

$=(4^{4})^{1/12}, (5^{3})^{1/12}, (6^{3})^{1/12}, (8^{4})^{1/12}$

$=(256)^{1/12}, (125)^{1/12}, (216)^{1/12}, (4096)^{1/12}$

$\therefore $ The smallest one is $\sqrt[4]{5}$

We first consider $\sqrt { 11 } -\sqrt { 10 }$ as follows:

(B) $\frac{p}{q}=\frac{\sqrt{32}-\sqrt{24}}{\sqrt{50}-\sqrt{48}}\times \frac{\sqrt{50}+\sqrt{48}}{\sqrt{50}+\sqrt{48}}$

$=\frac{(4\sqrt{2}-2\sqrt{6})(5\sqrt{2}+4\sqrt{3})}{2}$

$=(2\sqrt{2}-\sqrt{6})(5\sqrt{2}+4\sqrt{3})> 1$

$\therefore p> q$

(A) Given, $x=\sqrt{2}+1$ and $y=\sqrt{17}-\sqrt{2}$

$\dfrac{x}{y}=\dfrac{\sqrt{2}+1}{\sqrt{17}-\sqrt{2}}\times \dfrac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}$

$=\dfrac{(\sqrt{2}+1)(\sqrt{17}+\sqrt{2})}{17-2}$

$=\dfrac{(\sqrt{2}+1)(\sqrt{17}+\sqrt{2})}{15}< 1$

$\therefore x< y$

$ (0.5)^{2}=0.25$
$\sqrt{0.49}=0.7;$
$ \sqrt[3]{0.008}=\sqrt[3]{.2^3}=0.2$
$0.23$
Arranging in ascending order the numbers are $0.2< 0.23< 0.25< 0.7$
$ \therefore \sqrt[3]{0.008}=0.2$ is the least

LCM of $3, 2 = 6$
Given numbers are $ \sqrt[3]{2},\sqrt{3},\sqrt[3]{5}, 1.5$ i.e,
$ 2^{1/3},3^{1/2},5^{1/3},1.5$
$ \therefore $ Raising each number to power $6$, we get
$ (2^{1/3})^{6},(3^{1/2})^{6},(5^{1/3})^{6}, (1.5)^{6}$

$\displaystyle \sqrt{8}+\sqrt{5}=2.83+2.24=5.07$
$\displaystyle \sqrt{7}+\sqrt{6}=2.65+2.45=5.09$
$\displaystyle \sqrt{10}+\sqrt{13}=3.16+3.61=6.77$
$\displaystyle \sqrt{11}+\sqrt{12}=3.32+1.41=4.73$
$\displaystyle \therefore $ Smallest is $\displaystyle \sqrt{11}+\sqrt{2}$

$\sqrt[4]{10},\sqrt[3]{6},\sqrt{3}$
The order of the given irrational numbers are 2,3,4.
LCM of (2,3,4)=12
Now convert each irrational number as of order 12
$\sqrt[4]{10}=\sqrt[12]{10^3}=\sqrt[12]{1000}$
$\sqrt[3]{6}=\sqrt[12]{6^4}=\sqrt[12]{1296}$
$\sqrt{3}=\sqrt[12]{3^6}=\sqrt[12]{729}$
Hence, ascending order$\sqrt{3}<\sqrt[4]{10}<\sqrt[3]{6}$

Consider $4\sqrt { 18 }$ and factorize it as follows:

$(\sqrt{7}+\sqrt{10})$
$2.6457+3.1622=5.8079$
$(\sqrt{3}+\sqrt{19})$
$1.732+4.358=6.090$
Hence $(\sqrt{3}+\sqrt{19})$is greater.

LCM of $4, 6$ and $12 = 12$.
$ \therefore $ Raising each of the given number to power $12$, we have 
$ (3^{1/4})^{12},(10^{1/6})^{12},(25^{1/12})^{12}$
$= 3^{3},10^{2},25$
$= 27, 100, 25$
Arranging in descending order, the numbers are $ 100, 27, 25$
$\Rightarrow \sqrt[6]{10},\sqrt[4]{3},\sqrt[12]{25}$

$0.7+ \sqrt { 0.16 } = 0.7 +0.4 = 1.1$ 

Let us rewrite the given set of magnitudes $\sqrt [ 4 ]{ 5 } ,\sqrt [ 5 ]{ 4 } ,\sqrt { 4 },\sqrt {3}$ as follows:

$\sqrt{5} = 5^{1/2}$

$\sqrt[3]{11} = 11^{1/3}$

$2\sqrt[6]{3} = \sqrt[6]{12}= 12^{1/6}$

L.C.M of the denominators of the exponents is 12.

So,

$\sqrt{5} = 5^{\frac{1}{2}\times\frac{6}{6}} = \sqrt [12]{5^6}=\sqrt[12]{15625}$

$\sqrt[3]{11} = 11^{\frac{1}{3}\times\frac{4}{4}} = \sqrt [12]{11^4} =\sqrt[12]{14641}$

$2\sqrt[6]{3} = \sqrt[6]{12}= 12^{\frac{1}{6}\times\frac{2}{2}} = \sqrt[12]{12^2} =\sqrt[12]{144}$

Hence, the Ascending order is $2\sqrt[6]{3}, \sqrt[3]{11},\sqrt{5}$

$\displaystyle 6^\cfrac 14,2^\cfrac 12,4^\cfrac 13$
The surds are of the order $4, 2$ and $3$ respectively The L.C.M. is $12$ So, we change each surd of the order $12$
The terms now are $\displaystyle \left ( 6^{3} \right )^\cfrac {1}{12},\left ( 2^{6} \right )^\cfrac{1}{12}$ and $\displaystyle \left ( 4^{4} \right )^\cfrac{1}{12}$
$\displaystyle \Rightarrow \left ( 216 \right )^\cfrac{1}{12},\left ( 64 \right )^\cfrac{1}{12}$ and $\displaystyle \left ( 256 \right )^\cfrac{1}{12}$
$\displaystyle \therefore $ Ascending order is $\displaystyle \sqrt{2},\sqrt[4]{6}$ and $\displaystyle \sqrt[3]{4}$

$ \left(\dfrac{1}{2}\right)^{1/2} \; or \; \left(\dfrac{2}{3}\right)^{1/3}$

$= \left(\left(\dfrac{1}{2}\right)^{1/2}\right)^6 \; or \; \left(\left(\dfrac{2}{3}\right)^{1/3}\right)^6$

$= \left(\dfrac{1}{2}\right)^3 \; or \; \left(\dfrac{2}{3}\right)^2$

$= \left(\dfrac{1}{8}\right) \; or \; \left(\dfrac{4}{9}\right)$

= 0.125 or 0.44

Since, 0.44 is greater and so is $ \left(\dfrac{2}{3}\right)^{1/3}$

LCM of $3$ and $2$ is $6$.
Therefore, multiplying the index of all numbers by $6$.
${\sqrt [3]{2}}^6 = 2^\dfrac 63 = 2^2 = 4$

${\sqrt 3}^6 = 3^\dfrac 62 = 3^3 = 27$

${\sqrt 4}^6 = 4^ \dfrac 62 = 4^3 = 64$

${\sqrt 5}^6 = 5^{\dfrac 62} =  5^3 = 125$

$\therefore 125 > 64 > 27 > 4$

Option A: $\sqrt [ 3 ]{ -27 } ={ (-3) }^{ 3\times \frac { 1 }{ 3 }  }=-3$
Option B: $\sqrt { 2 } (3\sqrt { 2 } +2\sqrt { 8 } )=\sqrt { 2 } (3\sqrt { 2 } +4\sqrt { 2 } )=\sqrt { 2 } (7\sqrt { 2 } )=14$
Option C: $\dfrac { 3\sqrt { 18 }  }{ 2\sqrt { 6 }  } =\dfrac { 3\sqrt { 3 }  }{ 2 } $
Option D: $\sqrt { \dfrac { 1 }{ 2 }  } \sqrt { \dfrac { 25 }{ 2 }  } =\dfrac { 5 }{ 2 } $
Option E: $\dfrac { 2\sqrt { 5 }  }{ \sqrt { 45 }  } =\dfrac { 2\sqrt { 5 }  }{ 3\sqrt { 5 }  } =\dfrac { 2 }{ 3 } $
Therefore, all are rational except option $C$.